path: root/source/know/concept/second-quantization/index.md

---
title: "Second quantization"
sort_title: "Second quantization"
date: 2021-02-26
categories:
- Quantum mechanics
- Physics
layout: "concept"
---

The **second quantization** is a technique to deal with quantum systems
containing a large and/or variable number of identical particles.
Its exact formulation depends on
whether it is fermions or bosons that are being considered
(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)).

Regardless of whether the system is fermionic or bosonic,
the idea is to change basis to a set of many-particle wavefunctions
known as the **Fock states**, which are specific members of a **Fock space**
(a special kind of [Hilbert space](/know/concept/hilbert-space/))
with a well-defined number of particles.

For a set of $$N$$ single-particle energy eigenstates
$$\psi_k(x)$$ and $$N$$ identical particles $$x_k$$,
the Fock states are all the wavefunctions which contain $$n$$ particles,
for $$n$$ going from $$0$$ to $$N$$.

In this basis, we define the **particle creation operators**
and **particle annihilation operators**,
which respectively add/remove a particle to/from a given state.
In other words, these operators relate the Fock basis states
to one another, and are very useful.

The idea is to express states in such a way
that the fermionic/bosonic constraints are automatically satisfied,
and that the formulas look the same regardless of the number of particles.



## Fermions

Fermions need to obey the Pauli exclusion principle, so each state can only
contain one particle. In this case, the Fock states are given by:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            n &= 0:
            \qquad \Ket{0, 0, 0, ...}
            \\
            n &= 1:
            \qquad \Ket{1, 0, 0, ...} \quad \Ket{0, 1, 0, ...} \quad \Ket{0, 0, 1, ...} \quad \cdots
            \\
            n &= 2:
            \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots
            \\
            &\:\:\vdots \qquad \qquad \qquad \vdots
        \end{aligned}
    }
\end{aligned}$$

The notation $$\Ket{N_\alpha, N_\beta, ...}$$ is shorthand for
the appropriate [Slater determinants](/know/concept/slater-determinant/).
As an example, take $$\Ket{0, 1, 0, 1, 1}$$,
which contains three particles $$a$$, $$b$$ and $$c$$
in states 2, 4 and 5:

$$\begin{aligned}
    \Ket{0, 1, 0, 1, 1}
    = \Psi(x_a, x_b, x_c)
    = \frac{1}{\sqrt{3!}} \det\!
    \begin{bmatrix}
        \psi_2(x_a) & \psi_4(x_a) & \psi_5(x_a) \\
        \psi_2(x_b) & \psi_4(x_b) & \psi_5(x_b) \\
        \psi_2(x_c) & \psi_4(x_c) & \psi_5(x_c)
    \end{bmatrix}
\end{aligned}$$

The creation operator $$\hat{c}_\alpha^\dagger$$ and annihilation
operator $$\hat{c}_\alpha$$ are defined to live up to their name:
they create or destroy a particle in the state $$\psi_\alpha$$.
Formally, this means:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \hat{c}_\alpha^\dagger \Ket{...0_\alpha...}
            &= J_\alpha \Ket{...1_\alpha...}
            \\
            \hat{c}_\alpha \Ket{...1_\alpha...}
            &= J_\alpha \Ket{...0_\alpha...}
        \end{aligned}
    }
\end{aligned}$$

The factor $$J_\alpha$$ is sometimes known as the **Jordan-Wigner string**,
and is necessary here to enforce the fermionic antisymmetry,
when creating or destroying a particle in the $$\alpha$$th state:

$$\begin{aligned}
    J_\alpha
    = (-1)^{\sum_{j < \alpha} N_j}
\end{aligned}$$

So, for example, when creating a particle in state 4
of $$\Ket{0, 1, 1, 0, 1}$$, we get the following:

$$\begin{aligned}
    \hat{c}_4^\dagger \Ket{0, 1, 1, 0, 1}
    = (-1)^{0 + 1 + 1} \Ket{0, 1, 1, 1, 1}
\end{aligned}$$

The point of the Jordan-Wigner string
is that the order matters when applying the creation and annihilation operators,
so, for example:

$$\begin{aligned}
    \hat{c}_1^\dagger \hat{c}_2 \Ket{0, 1}
    &= \hat{c}_1^\dagger \Ket{0, 0}
    = \Ket{1, 0}
    \\
    \hat{c}_2 \hat{c}_1^\dagger \Ket{0, 1}
    &= \hat{c}_2 \Ket{1, 1}
    = - \Ket{1, 0}
\end{aligned}$$

In other words, $$\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger$$,
meaning that the anticommutator $$\{\hat{c}_2, \hat{c}_1^\dagger\} = 0$$.
You can verify for yourself that
the general anticommutators of these operators are given by:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \{\hat{c}_\alpha, \hat{c}_\beta\}
            &= 0
            \\
            \{\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger\}
            &= 0
            \\
            \{\hat{c}_\alpha, \hat{c}_\beta^\dagger\}
            &= \delta_{\alpha\beta}
        \end{aligned}
    }
\end{aligned}$$

Each single-particle state can only contain 0 or 1 fermions,
so these operators **quench** states that would violate this rule.
Note that these are *scalar* zeros:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \hat{c}_\alpha^\dagger \Ket{...1_\alpha...}
            &= 0
            \\
            \hat{c}_\alpha \Ket{...0_\alpha...}
            &= 0
        \end{aligned}
    }
\end{aligned}$$

Finally, as has already been suggested by the notation, they are each other's adjoint:

$$\begin{aligned}
    \matrixel{...1_\alpha...}{\hat{c}_\alpha^\dagger}{...0_\alpha...}
    = \matrixel{...0_\alpha...}{\hat{c}_\alpha}{...1_\alpha...}^{*}
\end{aligned}$$

Let us now use these operators to define the **number operator** $$\hat{N}_\alpha$$ as follows:

$$\begin{aligned}
    \boxed{
        \hat{N}_\alpha
        = \hat{c}_\alpha^\dagger \hat{c}_\alpha
    }
\end{aligned}$$

Its eigenvalue is the number of particles residing in state $$\psi_\alpha$$
(look at the hats):

$$\begin{aligned}
    \hat{N}_\alpha \Ket{... N_\alpha ...}
    = N_\alpha \Ket{... N_\alpha ...}
\end{aligned}$$



## Bosons

Bosons do not need to obey the Pauli exclusion principle, so multiple can occupy a single state.
The Fock states are therefore as follows:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            n &= 0:
            \qquad \Ket{0, 0, 0, ...}
            \\
            n &= 1:
            \qquad \Ket{1, 0, 0, ...} \quad \Ket{0, 1, 0, ...} \quad \Ket{0, 0, 1, ...} \quad \cdots
            \\
            n &= 2:
            \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots
            \\
            &\qquad\:\,\,
            \qquad \Ket{2, 0, 0, ...} \quad \Ket{0, 2, 0, ...} \quad \Ket{0, 0, 2, ...} \quad \cdots
            \\
            &\:\:\vdots \qquad \qquad \qquad \vdots
        \end{aligned}
    }
\end{aligned}$$

They must be symmetric under the exchange of two bosons.
To achieve this, the Fock states are represented by Slater *permanents*
rather than determinants.

The boson creation and annihilation operators $$\hat{c}_\alpha^\dagger$$ and
$$\hat{c}_\alpha$$ are straightforward:

$$\begin{gathered}
    \boxed{
        \begin{aligned}
            \hat{c}_\alpha^\dagger \Ket{... N_\alpha ...}
            &= \sqrt{N_\alpha + 1} \: \Ket{... (N_\alpha \!+\! 1) ...}
            \\
            \hat{c}_\alpha \Ket{... N_\alpha ...}
            &= \sqrt{N_\alpha} \: \Ket{... (N_\alpha \!-\! 1) ...}
        \end{aligned}
}\end{gathered}$$

Applying the annihilation operator $$\hat{c}_\alpha$$
when there are zero particles in $$\alpha$$ quenches the state:

$$\begin{aligned}
    \boxed{
        \hat{c}_\alpha \Ket{...0_\alpha...}
        = 0
    }
\end{aligned}$$

There is no Jordan-Wigner string, and therefore no sign change when commuting.
Consequently, these operators satisfy the following commutators:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            [\hat{c}_\alpha, \hat{c}_\beta]
            &= 0
            \\
            [\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger] 
            &= 0
            \\
            [\hat{c}_\alpha, \hat{c}_\beta^\dagger]
            &= \delta_{\alpha\beta}
        \end{aligned}
    }
\end{aligned}$$

The constant factors applied by $$\hat{c}_\alpha^\dagger$$ and $$\hat{c}_\alpha$$
ensure that $$\hat{N}_\alpha$$ keeps the same nice form:

$$\begin{aligned}
    \boxed{
        \hat{N}_\alpha
        = \hat{c}_\alpha^\dagger \hat{c}_\alpha
    }
\end{aligned}$$



## Operators

In the second quantization,
changing between different bases of single-particle states
is done in the usual way, where $$\alpha$$ and $$b$$ need not be in the same basis.
Note that $$\Ket{0}$$ is the zero-particle Fock state,
and $$\Ket{\alpha}$$ etc. are one-particle Fock states:

$$\begin{aligned}
    \hat{c}_b^\dagger \Ket{0}
    = \Ket{b}
    = \sum_{\alpha} \Ket{\alpha} \inprod{\alpha}{b}
    = \sum_{\alpha} \inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0}
\end{aligned}$$

With this, we define the **field operators**,
which create or destroy a particle at a position $$\vb{r}$$:

$$\begin{aligned}
    \boxed{
        \hat{\Psi}^\dagger(\vb{r})
        = \sum_{\alpha} \inprod{\alpha}{\vb{r}} \hat{c}_\alpha^\dagger
        \qquad \qquad
        \hat{\Psi}(\vb{r})
        = \sum_{\alpha} \inprod{\vb{r}}{\alpha} \hat{c}_\alpha
    }
\end{aligned}$$

By the same basis-changing principle,
any single-particle (non-interacting) operator $$\hat{V}$$ can be translated
to its second-quantized $$N$$-particle version as follows:

$$\begin{aligned}
    \hat{V}
    &= \sum_{\alpha, \beta} \ket{\alpha} \matrixel{\alpha}{\hat{V}}{\beta} \bra{\beta}
    = \sum_{\alpha, \beta} \ket{\hat{c}_\alpha^\dagger 0} \matrixel{\alpha}{\hat{V}}{\beta} \bra{\hat{c}_\beta^\dagger 0}
\end{aligned}$$

We take out the creation operators,
which allows us to generalize to multi-particle states:

$$\begin{aligned}
    \boxed{
        \hat{V}
        = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta
    }
\end{aligned}$$

Where the matrix element $$\matrixel{\alpha}{\hat{V}}{\beta}$$
is to be evaluated in the normal way:

$$\begin{aligned}
    \matrixel{\alpha}{\hat{V}}{\beta}
    = \int \psi_\alpha^*(\vb{r}) \: \hat{V}(\vb{r}) \: \psi_\beta(\vb{r}) \dd{\vb{r}}
\end{aligned}$$

In the same way, a two-particle interaction operator $$\hat{W}$$
can be rewritten in the form below.
Note the ordering of the operators' subscripts:

$$\begin{aligned}
    \boxed{
        \hat{W}
        = \sum_{\alpha, \beta, \gamma, \delta}
        W_{\alpha \beta \gamma \delta} \: \hat{c}_\alpha^\dagger \hat{c}_\beta^\dagger \hat{c}_\delta \hat{c}_\gamma
    }
\end{aligned}$$

Where the constant $$W_{\alpha \beta \gamma \delta}$$
is defined from the single-particle wavefunctions like so:

$$\begin{aligned}
    W_{\alpha \beta \gamma \delta}
    \equiv \iint \psi_\alpha^*(\vb{r}_1) \: \psi_\beta^*(\vb{r}_2)
    \: W(\vb{r}_1, \vb{r}_2) \: \psi_\gamma(\vb{r}_1)
    \: \psi_\delta(\vb{r}_2) \dd{\vb{r}_1} \dd{\vb{r}_2}
\end{aligned}$$



## References
1.  L.E. Ballentine,
    *Quantum mechanics: a modern development*, 2nd edition,
    World Scientific.